CODEWITHSUNDEEP
c++algorithmrecursionbig-ocombinations
Zhaonan
Zhaonan

Reputation: 939

What's time complexity of this algorithm for finding all combinations?

Combinations
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
For example, If n = 4 and k = 2, a solution is:

[
   [2, 4],
   [3, 4],
   [2, 3],
   [1, 2],
   [1, 3],
   [1, 4],
]

Personally I think,
time complexity = O(n^k), n and k are input.
Thank you for all help.
Finally, the time complexity = O(C(n,k) * k) = O((n!/(k! * (n - k)!)) * k), n and k is input,
Since, each time when we get a combination, we need copy subList list to one_rest, which is O(k), there is C(n, k) * k.
C++

#include <vector>
using namespace std;

class Solution {
public:
    vector<vector<int> > combine(int n, int k) {
        vector<vector<int>> list;

        // Input validation.
        if (n < k) return list;

        int start = 1;
        vector<int> subList;
        helper(n, k, start, list, subList);

        return list;
    }

    void helper(int n, int k, int start, 
                vector<vector<int>> &list, vector<int> &subList) {
        // Base case.
        if (subList.size() == k) {
            vector<int> one_rest(subList);
            list.push_back(one_rest);
            return;
        }
        if (start > n) return;

        for (int i = start; i <= n; i ++) {
            // Have a try.
            subList.push_back(i);

            // Do recursion.
            helper(n, k, i + 1, list, subList);

            // Roll back.
            subList.pop_back();
        }
    }
};

Upvotes: 32

Views: 44111

Answers (4)

Nuclearman
Nuclearman

Reputation: 5304

The complexity is O(C(n,k)) which is O(n choose k).

This ends up being equivalent to O(min(n^k, n^(n-k))).

Upvotes: 29

tsragravorogh
tsragravorogh

Reputation: 3153

I don't think it is O(n^k). Because think about it. Let's assume n=100 and k=2. According to your complexity it will 100 to the power 2. But if it is n=100 and k=10, it will be 100 to the power of 10. But if you think about it, there are far more combinations with n=100,k=2 than n=100,k=10. The complexity is actually is the actual formula: which is n!/(k!(n-k)!). And the complexity therefore will be O(n!/k!(n-k)!).

Upvotes: 4

Pradhan
Pradhan

Reputation: 16757

Since you are using lists, push_back and pop_back are O(1) operations. Also, you end up generating a valid combination exactly once. Thus, the complexity is O(n choose k).

Upvotes: 8

merlin2011
merlin2011

Reputation: 75575

The time complexity is equal to the number of combinations there are.

In this case it is n choose k.

Upvotes: 5

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